Problem: Find the limit as $x$ approaches positive infinity. $\lim_{x\to\infty}\dfrac{\sqrt{4x^2+4x}}{4x+1}=$
Solution: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the denominator is $x$, let's divide by $x$. In the numerator, let's divide by $\sqrt{x^2}$, since for positive values, $x=\sqrt{x^2}$. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{\sqrt{4x^2+4x}}{4x+1} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{\sqrt{4x^2+4x}}{\sqrt{x^2}}}{\dfrac{4x+1}{x}} \gray{\text{Divide sides by }x=\sqrt{x^2}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to\infty}\dfrac{\sqrt{\dfrac{4\cancel{x^2}}{\cancel{x^2}}+\dfrac{4\cancel x}{\cancel x\cdot x}}}{\dfrac{4\cancel{x}}{\cancel{x}}+\dfrac{1}{x}} \\\\ &=\lim_{x\to\infty}\dfrac{\sqrt{4+\dfrac{4}{x}}}{4+\dfrac{1}{x}} \\\\ &=\lim_{x\to\infty}\dfrac{\sqrt{4+0}}{4+0} \gray{\lim_{x\to\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{\sqrt{4}}{4} \\\\ &=\dfrac{2}{4} \\\\ &=\dfrac{1}{2} \end{aligned}$ In conclusion, $\lim_{x\to\infty}\dfrac{\sqrt{4x^2+4x}}{4x+1}=\dfrac{1}{2}$.